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Shearing in 2D Graphics
  • Difficulty Level : Hard
  • Last Updated : 01 Jul, 2020

Shearing deals with changing the shape and size of the 2D object along x-axis and y-axis. It is similar to sliding the layers in one direction to change the shape of the 2D object.It is an ideal technique to change the shape of an existing object in a two dimensional plane. In a two dimensional plane, the object size can be changed along X direction as well as Y direction.

x-Shear :
In x shear, the y co-ordinates remain the same but the x co-ordinates changes. If P(x, y) is the point then the new points will be P’(x’, y’) given as –
x'=x+Sh_x*y; y'=y

Matrix Form:
\begin{bmatrix}x'&y'\end{bmatrix}=\begin{bmatrix}x&y\end{bmatrix}.\begin{bmatrix}1&0\\Sh_x&1\end{bmatrix}
y-Shear :
In y shear, the x co-ordinates remain the same but the y co-ordinates changes. If P(x, y) is the point then the new points will be P’(x’, y’) given as –
x'=x; y'=y+Sh_y*x



Matrix Form:
\begin{bmatrix}x'&y'\end{bmatrix}=\begin{bmatrix}x&y\end{bmatrix}.\begin{bmatrix}1&Sh_y\\0&1\end{bmatrix}

x-y Shear :
In x-y shear, both the x and y co-ordinates changes. If P(x, y) is the point then the new points will be P’(x’, y’) given as –
x'= x+Sh_x*y; y'=y+Sh_y*x

Matrix Form:
\begin{bmatrix}x'&y'\end{bmatrix}=\begin{bmatrix}x&y\end{bmatrix}.\begin{bmatrix}1&Sh_y\\Sh_x&1\end{bmatrix}

Example :
Given a triangle with points (1, 1), (0, 0) and (1, 0).
Find out the new coordinates of the object along x-axis, y-axis, xy-axis.
(Applying shear parameter 4 on X-axis and 1 on Y-axis.).

Explanation –

Given,
Old corner coordinates of the triangle = A (1, 1), B(0, 0), C(1, 0)
Shearing parameter along X-axis (Shx) = 4
Shearing parameter along Y-axis (Shy) = 1

Along x-axis:
A'=(1+4*1, 1)=(5, 1)
B'=(0+4*0, 0)=(0, 0)
C'=(1+4*0, 0)=(1, 0)

Along y-axis:
A''=(1, 1+1*1)=(1, 2)
B''=(0, 0+1*0)=(0, 0)
C''=(1, 0+1*1)=(1, 1)

Along xy-axis:
A'''=(1+4*1, 1+1*1)=(5, 2)
B'''=(0+4*0, 0+1*0)=(0, 0)
C'''=(1+4*0, 0+1*1)=(1, 1) 

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